## Wanted: Elegant proof of sin(x)/x limit

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Qaanol
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### Wanted: Elegant proof of sin(x)/x limit

Is there an elegant, elementary, preferably pictorial proof that sin(x)/x → 1 as x → 0?

I know of a simple way to show that (1-cos(x))/x → 0, but I have not yet found a correspondingly transparent illustration for sin(x)/x.

The most promising ones I’ve seen use a trio of nested shapes: a triangle of base 1 and height sin(x), inside a sector of the unit circle with arc length x, inside a right triangle of base 1 and height tan(x). In the version using arc length directly, the difficult step is to show that tan(x) > x on (0, π/2), and in the version using areas the difficult step is to show that a sector of the unit circle has area equal to half its arc length.

So is there an easy way to prove either of the statements “tan(x) > x on (0, π/2)” or “Asector = θr²/2”, or perhaps a better approach to solving the sin(x)/x limit altogether?
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### Re: Wanted: Elegant proof of sin(x)/x limit

If you can use area of circle = pi r^2, then you can show the area of a sector as a ratio of the circle area: pi r^2 * (theta/2pi) = theta r^2 /2

Qaanol
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### Re: Wanted: Elegant proof of sin(x)/x limit

That shunts the difficulty over to showing the ratio between a circle’s area and perimeter is r/2.

Since a straight line is the shortest distance between two points (I’m taking this as axiom for now), a regular polygon inscribed in a circle has smaller perimeter than the circle does. And since containment is a partial order on areas, a regular polygon circumscribed about a circle has greater area than the circle does. Comparing the edges of the inner polygon to the matching triangular slices of the outer polygon and letting the edge counts explode, it follows that the perimeter of a circle is at least r/2 times its circumference.

But how can we put an upper bound on that ratio?

For instance, if we could show that the outer polygon’s perimeter is longer than the circle’s, we would be done. But that is morally equivalent to showing that tan(x) > x on (0, π/2). From what I’ve read, Archimedes used the fact that both the circle and the circumscribing polygon are convex, but I have yet to find a good proof that if one convex shape contains another then the perimeter of the larger exceeds the perimeter of the smaller.
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jestingrabbit
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### Re: Wanted: Elegant proof of sin(x)/x limit

Qaanol wrote:“tan(x) > x on (0, π/2)”

BC has length tanx, and BD has length x. I claim |BD| < |BE|+|ED| < |BE| + |EC| = |BC| with the first inequality verified by good sense and the second by EDC being a right angle, and the hypotenuse being the longest side of a right triangle.
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Qaanol
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### Re: Wanted: Elegant proof of sin(x)/x limit

jestingrabbit wrote:I claim |BD| < |BE|+|ED|
<snip>
verified by good sense

That’s the part I’m struggling to prove cleanly.
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quintopia
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### Re: Wanted: Elegant proof of sin(x)/x limit

It's not obvious to me how the "good sense" step is made rigorous using only elementary ideas. Indeed it appears to suffer from Qaanol's complaint:
Qaanol wrote:I have yet to find a good proof that if one convex shape contains another then the perimeter of the larger exceeds the perimeter of the smaller.

My best guess as to how to patch it up is to simply repeat jestingrabbit's argument recursively on the triangles AEB and ADE by adding a point F where AE crosses the circle and the corresponding tangent. Add G where the new tangent crosses BE and the same argument shows that B->G->F is shorter than BE. Then recursively repeat it again ad infinitum until you have the actual arc of the circle in the limit.

Qaanol
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### Re: Wanted: Elegant proof of sin(x)/x limit

Going that route, repeatedly “slicing off” the corners of the outer piecewise-linear path, will indeed bring it arbitrarily close to the circle. But then we’re treading in the territory of actually defining what “arc length of a curve” means, which is rather heavy machinery for this problem. I don’t want to deal with Jordan curves and fractal boundaries and so forth.

It is worth noting that the limit of (1−cos(x))/x can be found without any such definition of arc length, just using the fact that a straight line is the shortest distance between two point.

Perhaps we can get the result about nested convex shapes by lifting up a dimension to have a cone inside a not-necessarily-circular cylinder. I can’t quite get it to work though, since any “pointy” parts of the inner shape may well end up having a locally reduced perimeter when the cone meets the cylinder wall.

That’s not a problem when the inner shape is a circle, of course, so maybe something in this vein can be achieved for the specific case at hand.

The image of “wrapping” a sector of a circle into a cone, and then “unwrapping” the contour rings of the cone at constant height, strikes me as poignant. The result is to transform the circular sector into a triangle of height r and base equal to the arc length, but I don’t know if that approach can easily be demonstrated to preserve area.

Heck, we haven’t even put any upper bound on the circumference of a circle yet. We might need the nested-convex-shapes theorem just to show it’s finite!
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quintopia
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### Re: Wanted: Elegant proof of sin(x)/x limit

Qaanol wrote:That shunts the difficulty over to showing the ratio between a circle’s area and perimeter is r/2.

Just as a matter of course, what are your qualms with this proof?

http://www.ams.org/samplings/feature-column/fc-2012-02

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### Re: Wanted: Elegant proof of sin(x)/x limit

Qaanol wrote:
jestingrabbit wrote:I claim |BD| < |BE|+|ED|
<snip>
verified by good sense

That’s the part I’m struggling to prove cleanly.

This is the Triangle Inequality and that wiki page has a short euclidean proof.

quintopia
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### Re: Wanted: Elegant proof of sin(x)/x limit

It is not the triangle inequality. Triangles do not have arcs for sides. But the link I posted above gives some ideas about how one might approach the issue. Not a complete proof, but a good starting point.

jestingrabbit
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### Re: Wanted: Elegant proof of sin(x)/x limit

Yeah, I apologise for the notation there. BD is not the interval, but the section of the circle.

I think that if you want a rigorous proof, you'll need calculus because that is the only way to remove these sorts of niggles, but if you want an elegant proof you are going to have these sorts of appeals to basic intuition.
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Qaanol
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### Re: Wanted: Elegant proof of sin(x)/x limit

quintopia wrote:Just as a matter of course, what are your qualms with this proof?

http://www.ams.org/samplings/feature-column/fc-2012-02

Well, one thing that jumps out is that no proof of the nested-convex-shapes fact is given for curves. Yes, it is quite readily shown for polygons, but not necessarily so for circles.

And for the triangle with area equal to the circle, the construction shows how to find its side lengths, but to say the base of the triangle equals the circumference of the circle is to define arc length as a limit of piecewise-linear segments. Yes it works, but proving that the resulting arc length is well-defined (ie. the same result no matter what sequence of approximating polygons is used) should not be considered trivial.

For the specific case of two tangent lines outside a circle (Jestingrabbit’s BE and ED) however, I am hoping there is a direct proof that the arc is shorter than the two lines.
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Derek
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### Re: Wanted: Elegant proof of sin(x)/x limit

Can we accept that in the diagram, the path outside the circle must be longer than the path inside the circle? I'm not sure what the best approach for a formal but elementary proof would be, but it's intuitively clear that a path outside of a convex curve cannot be shorter than that curve. I think I could construct a formal proof using calculus and polar integration of arclength, but that's probably not elementary. But if we accept this, then we have that 2*tan(x) > 2x, therefore tax(x) > x. (In the diagram I've used 90 degree angels for convenience of drawing, but the property holds for any x.)

If you don't accept my claim, we've at least established the additional sufficient condition of proving that curves outside of a convex curve must be longer than the convex curve.

EDIT: I realize now that this is basically the same thing that jestingrabbit said.
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jestingrabbit
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### Re: Wanted: Elegant proof of sin(x)/x limit

Just to lay out some of the more obvious things that we do actually know. We definitely have that tan is monotonically increasing and unbounded on the relevant interval. Further, I believe that the tan angle sum is just a consequence of basic trig stuff, no limits needed. So, from this, we can conclude that there is a greatest t such that t = tan(t). I think rubbing these two things together in the right way should lead to a contradiction of some sort, but its not going to be elegant, that is for sure.
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kubit
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### Re: Wanted: Elegant proof of sin(x)/x limit

Is the some reason we don't want to apply L'Hospital's rule?

Nicias
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### Re: Wanted: Elegant proof of sin(x)/x limit

kubit wrote:Is the some reason we don't want to apply L'Hospital's rule?

L'Hospital's rule requires the derivative of sin(x), which requires knowledge of this limit.

Using L'H in this case is begging the question.

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### Re: Wanted: Elegant proof of sin(x)/x limit

I don't think this can be done without at least the beginnings of a definition of lengths of curves. But we don't need full-blown calculus. It should suffice to take as an axiom that in the limit of zero length, a curve is approximated at a point by its tangent line at that point. With that, it becomes trivial to show that sin(x)=x in the limit as x approaches zero, so sin(x)/x becomes x/x=1. That does smell a bit like cheating, but I think any theory of curve lengths at some point has to just define the approximation by tangents to be correct.

kubit
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### Re: Wanted: Elegant proof of sin(x)/x limit

Nicias wrote:
kubit wrote:Is the some reason we don't want to apply L'Hospital's rule?

L'Hospital's rule requires the derivative of sin(x), which requires knowledge of this limit.

Using L'H in this case is begging the question.

Doh.. such a brain fart from me. How about using sin(x) series expansion?

Demki
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### Re: Wanted: Elegant proof of sin(x)/x limit

kubit wrote:
Nicias wrote:
kubit wrote:Is the some reason we don't want to apply L'Hospital's rule?

L'Hospital's rule requires the derivative of sin(x), which requires knowledge of this limit.

Using L'H in this case is begging the question.

Doh.. such a brain fart from me. How about using sin(x) series expansion?

That, sadly(or not!), requires the derivative of sin(x) too.

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### Re: Wanted: Elegant proof of sin(x)/x limit

jestingrabbit wrote:Just to lay out some of the more obvious things that we do actually know. We definitely have that tan is monotonically increasing and unbounded on the relevant interval. Further, I believe that the tan angle sum is just a consequence of basic trig stuff, no limits needed.

Yes, both of those things can be shown without heavy machinery.

jestingrabbit wrote:So, from this, we can conclude that there is a greatest t such that t = tan(t). I think rubbing these two things together in the right way should lead to a contradiction of some sort, but its not going to be elegant, that is for sure.

In order to conclude from those premises that there is a greatest fixed point of tan(x) on [0, π/2), we would also need to prove that “π/2” (which I am using to denote the arc length of a quarter circle) is itself finite. Put another way, we need an upper bound on the circumference of a circle first.

It is easy enough to put a lower bound on circumference, because a straight line is the shortest path between endpoints. And it is easy enough to put both upper and lower bounds on area, because containment puts a partial order on area. But I do not see an obvious way to put an upper bound on the length of a curve.

Hmm, if we assume that a circle has finite circumference which grows with radius (we don’t even need that it’s proportional, just monotone!), then we can develop some extremely rudimentary “calculus”. We’ll need the fact that the area of a circle is proportional to its radius squared, which is easy enough to show with triangles. We also need one more fact, which I will call the…

Hat brim lemma: In a circular top hat of height h, if the brim extends out from the hat by that same distance h, then the brim has larger area than the vertical “stovepipe” wall of the hat.

Proof: Remove the flat top of the hat and unroll the wall-and-brim. The outer part of the brim shrinks during this process, while the vertical wall keeps its area. When they are straightened, both pieces become rectangles of equal width (h) and length (the inner circumference), so they end up with equal areas. The brim shrank and the wall didn’t, so the brim was larger at the outset.

Now we can compare a circle of radius r with one of radius (r+h). Call their areas A and B, respectively, and let c be the circumference of the smaller circle. By the hat brim lemma we have B > A+hc. But area is proportional to radius squared, so we also know that A/r2 = B/(r+h)2, which yields B = A(r+h)2/r2.

Plugging that into the previous inequality gives A(r+h)2/r2 > A+hc. Some algebra (ugh, is there a good way to show this geometrically?) yields A(2hr+h2)/r2 > hc, and then A(2 + h/r) > rc.

For a fixed radius r, we can take h as small as we like and the inequality still holds. Since A is a constant, it follows that 2A cannot be less than rc (otherwise a small enough h would break the inequality), so we must have A ≥ rc/2.

The same argument holds for any sector of a circle (just slice the top hat accordingly) which means Asector ≥ rs/2. We could do the same thing but with the wall outside the brim (so the brim expands) to show that A ≤ rx/2, and thus A = rx/2 as well as Asector = rs/2, but for now just having the upper bound suffices.

Looking at the unit circle (in the first quadrant), by containment of areas we have tan(x)/2 > Asector. With s=x and r=1 we get Asector ≥ x/2, hence tan(x)/2 > x/2 or simply tan(x) > x. Taking reciprocals and multiplying by sin(x) gives cos(x) < sin(x)/x. The upper bound of 1 is easy to get, as is the fact that cos(x) → 1. Symmetry gives us the same results in the third quadrant, which shows that sin(x)/x → 1 if we accept the hat brim lemma (including its assumption that circles have finite circumference).

Of course, my justification for the hat brim lemma is not exactly rigorous. It appeals to intuition, and is boosted by the fact that circles are radially-symmetric so the same contraction of the brim happens in all directions, but it still talks about bending surfaces in three dimensions.

Can we provide a stronger (preferably two-dimensional) foundation for the hat-brim lemma, which shows that the area of an annulus is at least its radial width times its inner circumference?
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Derek
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### Re: Wanted: Elegant proof of sin(x)/x limit

Demki wrote:That, sadly(or not!), requires the derivative of sin(x) too.

Not if you define sin(x) by it's series expansion, which is not as crazy as it sounds.

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### Re: Wanted: Elegant proof of sin(x)/x limit

Derek wrote:
Demki wrote:That, sadly(or not!), requires the derivative of sin(x) too.

Not if you define sin(x) by it's series expansion, which is not as crazy as it sounds.

You could also define sine as the solution to the IVP: f''(x)=-f(x), f(0)=0, f'(0)=1. Which gives the limit immediately.

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### Re: Wanted: Elegant proof of sin(x)/x limit

Nicias: now prove that that sine is also the right triangle side ratio.

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### Re: Wanted: Elegant proof of sin(x)/x limit

quintopia wrote:Nicias: now prove that that sine is also the right triangle side ratio.

Simple, just apply Euler's formula (which can be proven with only the power series definition or the differential equation definition) and the geometric interpretation of complex numbers.

EDIT: You'll have to show that sin^2 + cos^2 = 1 from the power series only, but that should be possible with some careful algebra.

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### Re: Wanted: Elegant proof of sin(x)/x limit

Another approach: Take a straight-edge, pin one end, and rotate it. The point at distance 1 on this ruler moves at constant speed and traces out the unit circle.

Now consider another point that begins at distance 1 and moves outward along the straight-edge to trace the tangent line x=1. As the ruler rotates, this point has radial and tangential velocity components. Its tangential speed is at least as fast as the first point’s, because this one is further out on the same rotating ruler. Meanwhile its radial velocity is perpendicular to its tangential velocity.

If we accept that components of instantaneous velocity combine via the Pythagorean theorem, then it follows that the speed of the outer point is at least its tangential speed, which is itself at least as fast as the inner point. So the outer point is never slower than the inner.

When we rotate the straight-edge through an angle θ that is less than a quarter circle, the inner point moves distance θ and the outer point moves distance tan(θ). The faster point must have gone further, so tan(θ) ≥ θ.

We might also be able to do something with the fact that a circle has the least perimeter of all shapes for its area, but that is notoriously difficult to prove. It’s straightforward to show that if such a minimal shape exists then it must be a circle, but the existence step is tricky.
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Derek
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### Re: Wanted: Elegant proof of sin(x)/x limit

That's essentially what I had in mind when I said, "I think I could construct a formal proof using calculus and polar integration of arclength".

We might also be able to do something with the fact that a circle has the least perimeter of all shapes for its area, but that is notoriously difficult to prove. It’s straightforward to show that if such a minimal shape exists then it must be a circle, but the existence step is tricky.

Out of curiosity, what's the straightforward proof here?

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### Re: Wanted: Elegant proof of sin(x)/x limit

What about considering a circle as the limit of a regular n-gon as n->infinity?

Consider a regular n-gon:
It can be broken up into n congruent isosceles triangles (radially symmetric) about the center. Let the two equal length edges of each triangle to be 1. Let the apex angle be x and each of the non apex angles be y.
The sine of the apex angle is just the altitude dropped down from one of the non-apex angles. Let the length of that altitude be s.
The perimeter of the polygon is given by n*t, where t is the length of the edge opposite to the apex angle in one of the isosceles triangles.

sine(y) = sine(90-0.5x) = s/t, using the definition of sine.
Since all the triangles are congruent and are arranged with radial symmetry, x = 360/n
Clearly as n->infinity s/t -> 1

s = sin(x) by construction. Notice t is 1/n of the polygon perimeter (P) and x is 1/n of 2pi (rad).
If we can establish P->2pi as n->infinity, then we're done.

Consider a unit circle inscribed in the the regular n-gon. If you were to calculate the circle's perimeter using arc-length calculus, you would abstractly construct a sequence of interconnected segments that form the n-gon (arguing using symmetry), then sum the lengths of all the segments.
Thus, P -> perimeter of a unit circle as n->infinity.
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### Re: Wanted: Elegant proof of sin(x)/x limit

Cradarc wrote:Consider a unit circle inscribed in the the regular n-gon. If you were to calculate the circle's perimeter using arc-length calculus, you would abstractly construct a sequence of interconnected segments that form the n-gon (arguing using symmetry), then sum the lengths of all the segments.
Thus, P -> perimeter of a unit circle as n->infinity.

This is the part that doesn't seem straightforward or rigorous under the criteria of the current investigation. Qaanol doesn't want to consider whether a regular n-gon approaches a circle as n goes to infinity.

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### Re: Wanted: Elegant proof of sin(x)/x limit

Re: the Hatbrim Lemma - If appealing to notions like roll up and unroll is rigorous enough, you should be able to do a 2D analogue of the hatbrim lemma with nothing more than some constructions + rolling/unrolling. I don't think there's any way to fully rigorously construct a circumference equal to a line segment or vice versa, which it seems you'd need to do that without the top hat construction.
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### Re: Wanted: Elegant proof of sin(x)/x limit

Derek wrote:
quintopia wrote:Nicias: now prove that that sine is also the right triangle side ratio.

Simple, just apply Euler's formula (which can be proven with only the power series definition or the differential equation definition) and the geometric interpretation of complex numbers.

EDIT: You'll have to show that sin^2 + cos^2 = 1 from the power series only, but that should be possible with some careful algebra.

Actually you get the pythagorean identity immediately. Cosine is defined by f'' = -f, f'(0)=0, f(0)=1. So you immediately get that the derivative of sine is cosine and the derivative of cosine is negative sine.

Then sin^2(0)+cos^2(0)=1 while d/dx (sin^2(x)+cos^2(x)) = 2 sin(x) cosIx) + 2 cos(x) (-sin(x)) =0. Thus sin^2(x)+cos^2(x) is a constant, namely 1.

Euler's formula follows from the power series expansion immediately. I don't see how it gets you trigonometry.

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### Re: Wanted: Elegant proof of sin(x)/x limit

Its certainly the case that you can either develop this stuff from a calculus stance, where, for instance, you define pi/2 to be the first positive root of cos, or there's another line of development, where you start with something like a classical geometric perspective. If you start with calculus, then the result is clear, so we definitely need to commence our reasoning in a classical framework.

Ultimately though, if you want the rigor that Qaanol wants, you pretty much need to develop a lot of the calculus framework it seems to me. I mean this

Qaanol wrote:Another approach: Take a straight-edge, pin one end, and rotate it. The point at distance 1 on this ruler moves at constant speed and traces out the unit circle.

Now consider another point that begins at distance 1 and moves outward along the straight-edge to trace the tangent line x=1. As the ruler rotates, this point has radial and tangential velocity components. Its tangential speed is at least as fast as the first point’s, because this one is further out on the same rotating ruler. Meanwhile its radial velocity is perpendicular to its tangential velocity.

If we accept that components of instantaneous velocity combine via the Pythagorean theorem, then it follows that the speed of the outer point is at least its tangential speed, which is itself at least as fast as the inner point. So the outer point is never slower than the inner.

When we rotate the straight-edge through an angle θ that is less than a quarter circle, the inner point moves distance θ and the outer point moves distance tan(θ). The faster point must have gone further, so tan(θ) ≥ θ.

Is really only rigorous if you've got a fair amount of calculus. (I mean, there was the suggestion earlier in the thread that we don't know whether pi/2 is finite, so if we want rigor on that, we need rigor here too surely).

So I think that's where we find ourselves. Either its clear, or it needs a little geometric intuition ie one convex form contains another, then the contained form has a lesser perimeter.
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### Re: Wanted: Elegant proof of sin(x)/x limit

If you want more rigor, you will need to provide a "rigorous" definition of pi. You can prove something closely related to the circle without invoking the definition of pi.
I seem to recall pi is derived by looking at the convergence of the perimeter of regular n-gons as n->infinity.
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### Re: Wanted: Elegant proof of sin(x)/x limit

Carlington wrote:I don't think there's any way to fully rigorously construct a circumference equal to a line segment or vice versa, which it seems you'd need to do that without the top hat construction.

There's not. As I recall, this is equivalent to squaring the circle.

Nicias wrote:Euler's formula follows from the power series expansion immediately. I don't see how it gets you trigonometry.

Hmm. If we're taking this approach, I believe it would follow from the definition of radians. Consider some point x + y*i. This can be represented using Euler's Formula as sqrt(x^2 + y^2)e^i*a for some value a. Define a as the angle between (0, x + y*i) and (0, 1). If pi is defined with respect to the roots of sin or cos, then it naturally follows that 2pi radians makes a complete cycle.

Proving that radians are the ratio of arclength to radius however requires a rigorous definition of arclength, which is what we've been struggling with without bringing in more calculus. However, I think the rest of trigonometry can be recovered from the definitions I've given above.

I think the big thing I've taken away from this thread is that geometry, at least circular geometry, can not be well founded without calculus. Even the classical definition of pi (whether by circumference or area, though it seems like area is more rigorous) relies on taking a limit. So I don't think there is a rigorous answer to Qaanol's question that doesn't involve, at some level, using limits to define arclength or defining sin with respect to a power series (which is also limit) or the solution to a differential equation.

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### Re: Wanted: Elegant proof of sin(x)/x limit

Derek wrote:
We might also be able to do something with the fact that a circle has the least perimeter of all shapes for its area, but that is notoriously difficult to prove. It’s straightforward to show that if such a minimal shape exists then it must be a circle, but the existence step is tricky.

Out of curiosity, what's the straightforward proof here?

Steiner’s method.

So if we accept that there is some shape which maximizes area for its perimeter, then that shape is a circle. Any shape that contains a circle has larger area, so it would therefore have longer perimeter. This is enough to guarantee Jestingrabbit’s “good sense” claim, and hence tan(x) > x on (0, π/2).

The problem is, Steiner’s method says “Given a shape which is a not a circle, here is a way to make a shape with larger area and the same perimeter.”

For comparison, consider the statement, “Given a positive integer which is not one, squaring is a way to make a larger positive integer.”

We cannot conclude “One is the largest positive integer” until we know that such a maximal element exists.

Derek wrote:I think the big thing I've taken away from this thread is that geometry, at least circular geometry, can not be well founded without calculus. Even the classical definition of pi (whether by circumference or area, though it seems like area is more rigorous) relies on taking a limit. So I don't think there is a rigorous answer to Qaanol's question that doesn't involve, at some level, using limits to define arclength or defining sin with respect to a power series (which is also limit) or the solution to a differential equation.

I am fine with using limits, as long as it is the limit of a specific sequence of quantities.

For example, every polygon inside a unit circle has smaller area than that circle. For any area A less than π (meaning the number 3.14159…) there is a polygon with area greater than A inside the unit circle (proof by inscribed regular 3·2n-gons). Therefore the unit circle has area at least π.

Also, every polygon outside the unit circle has area greater than that circle. For any area B greater than π, there is a polygon with area less than B outside the unit circle. So the unit circle has area at most π, and therefore equal to π.

That type of limit is fine.

Similarly, every polygon inscribed in the unit circle has smaller perimeter than that circle. For any length L less than 2π, there is a polygon with perimeter greater than L inscribed in the unit circle. Therefore the unit circle has perimeter at least 2π.

If we knew that circumscribed polygons have greater perimeter than the circle, then we could show that the upper bound is also 2π. But so far we don’t.

We are very close, and any of the following would get us to the sin(x)/x limit:
i. Archimedes’ claim that a convex shape has lesser perimeter than any convex shape outside it.
ii. A circle has lesser perimeter than any circumscribing polygon.
iii. There exists a shape which maximizes area for a given perimeter.
iv. The area of an annulus of thickness h and inner circumference c is at least hc (and circles have finite circumference).
v. Instantaneous velocity is a Pythagorean sum of orthogonal components.
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### Re: Wanted: Elegant proof of sin(x)/x limit

Qaanol wrote:iv. The area of an annulus of thickness h and inner circumference c is at least hc (and circles have finite circumference).

C = 2*pi*r
A = pi*r^2
ΔA/Δr = (pi*((r+Δr)^2 - r^2))/Δr = pi*(2r+Δr) = C + pi*Δr
ΔA = area of annulus with thickness Δr.

So we have ΔA = Δr(C + pi*Δr) = ΔrC + pi*(Δr)^2
Clearly pi*(Δr)^2 >= 0.
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quintopia
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### Re: Wanted: Elegant proof of sin(x)/x limit

Cradarc: That first step is suspicious. How did you determine the constant scaling factor between a circle's radius and circumference?

Qaanol
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### Re: Wanted: Elegant proof of sin(x)/x limit

All right, I think I’ve gotten to the point where I’m willing to accept Archimedes’ statement that a convex shape has longer perimeter than any convex shape inside it, justified by etymological intuition.

The word “line” comes from Latin for “linen” (and “straight” from Old English for “stretched”) so it seems reasonable to postulate the length of a path as the length of a (perfect one-dimensional bendable-but-unstretchable) string along that path.

A piece of string along an outer convex shape will be slack around any inner convex shape, so if it is pulled tight around the inner shape there will be extra string left over. I’ll call this the “belt tightening lemma”.

Similarly, we can take a string along tan(x) and fold it over as in Jestingrabbit’s illustration. It will extend beyond end of the arc, and still have slack to take up. When we pull it tight around the circle it will thus extend even further beyond the end of the arc, so tan(x) > x.
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PM 2Ring
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### Re: Wanted: Elegant proof of sin(x)/x limit

FWIW, the Wikipedia article on the Fermat point (the point such that the total distance from the three vertices of the triangle to the point is the minimum possible) invokes without proof "the dogleg rule"
Wikipedia wrote:which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon.

Sadly, I can't find any other references to the dogleg rule.

Sizik
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### Re: Wanted: Elegant proof of sin(x)/x limit

First google result I get is from GeoGebra, which has it like this:
Attachments
she/they
gmalivuk wrote:
King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.

Qaanol
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### Re: Wanted: Elegant proof of sin(x)/x limit

Oh yes, it is quite easily shown that if two convex polygons are nested, then the outer has greater perimeter than the inner. Just go around the inner one leg by leg, extend each leg in both directions until it hits the outer polygon, and replace the part of the outer polygon between those points with the new line, so the outer still contains the inner.

Since lines make shortest distances, this new outer polygon has smaller perimeter than the old one. Repeat for each leg of the inner polygon, and in a finite number of length-reducing steps we’ve transformed the outer polygon into the inner one. This actually proves the stronger claim “a convex polygon has lesser perimeter than any shape that contains it”. (If the outer shape crosses an extended leg infinitely many times, just pick one crossing on each side however you like.)
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